Question Video: Finding the Value of an Expression Using the Distributive Property of Cross Product Mathematics

If ๐€ and ๐ are unit vectors and ๐œƒ the measure of the angle between them, find |(๐€ โˆ’ ๐) ร— (๐€ + ๐)|.

04:55

Video Transcript

If ๐€ and ๐ are unit vectors and ๐œƒ the measure of the angle between them, find the magnitude of the cross product between the vector ๐€ minus the vector ๐ and the vector ๐€ plus the vector ๐.

In this question, weโ€™re given two unit vectors ๐€ and ๐, and weโ€™re also told that ๐œƒ is the measure of the angle between these two unit vectors. And we need to use this information to simplify an expression, the magnitude of the cross product between the difference between these two vectors and the sum of these two vectors.

And thereโ€™s a few different ways we could approach this problem. For example, we could approach this problem graphically. However, weโ€™re going to approach this problem algebraically. And to approach this problem algebraically, we need to start by noting weโ€™re taking the cross product of the difference of two vectors and the sum of two vectors. We can then recall that the cross product is distributive over both vector addition and vector subtraction. This means we can distribute this cross product in a very similar way we would if this was two binomials.

And there is one key difference that we want to show. So letโ€™s start with the first term in each set of parentheses. We get the cross product between ๐€ and ๐€. Then we move on to the outer terms in each bracket. We get the cross product between vectors ๐€ and ๐. Next, we take the cross product of the inner two vectors. In this case, thatโ€™s negative ๐ and ๐€. And finally, we take the cross product of the last vectors. Thatโ€™s the cross product between negative ๐ and ๐. Then we take the magnitude of this entire vector expression. And now we can highlight the key difference we have in this distribution.

The vector cross product is not commutative. Usually, we would be able to cancel the middle two terms in this expansion. However, negative ๐ cross ๐€ is not equal to negative ๐€ cross ๐. In fact, if we switch the order in our cross products, we actually multiply by negative one. For any vectors ๐ฎ and ๐ฏ, negative ๐ฎ cross ๐ฏ is equal to ๐ฏ cross ๐ฎ. So we can use this result to replace negative ๐ cross ๐€ with ๐€ cross ๐.

And this is not the only piece of simplification we can do. We know weโ€™re taking the cross product between vector ๐€ and itself. And weโ€™re also taking the cross product between negative ๐ and ๐. And thereโ€™s a few different ways we can simplify these two terms. One way is to use the fact that for any vector ๐ฎ, cross product of it with itself will give the zero vector. We can apply this to ๐€ cross ๐€. And we can also apply this to our final term by taking the vector of negative one outside of our cross product. This would then give us that the first and final term in this expression are both the zero vector.

Itโ€™s worth noting this wasnโ€™t the only way of simplifying this expression. We could have also used a very similar fact. And thatโ€™s the fact that the cross product between any two parallel vectors is the zero vector. Using either method, we were able to simplify this expression to the magnitude of the cross product between ๐€ and ๐ added to the cross product between vectors ๐€ and ๐. And of course, weโ€™re adding the same term to itself. So we can rewrite this. Itโ€™s equal to the magnitude of two times the cross product between vectors ๐€ and ๐.

And now to simplify this expression any further, weโ€™re going to need to use the definition of the cross product. We recall for any vectors ๐ฎ and ๐ฏ, the cross product between ๐ฎ and ๐ฏ is equal to the magnitude of vector ๐ฎ times the magnitude of vector ๐ฏ multiplied by the sine of the angle between the two vectors ๐œ‘ times a vector ๐ฐ, where ๐ฐ is a unit vector perpendicular to ๐ฎ and ๐ฏ. And its direction is determined by the right-hand rule.

But we donโ€™t need to worry about this unit vector in this case, since weโ€™re taking the magnitude of this expression. And to see this, letโ€™s take the magnitude of both sides of this equation. The magnitude of a product is equal to the product of the magnitudes. So the right-hand side of this equation simplifies to give us the magnitude of ๐ฎ times the magnitude of ๐ฏ multiplied by the magnitude of sin ๐œ‘ times the magnitude of ๐ฐ. But ๐ฐ is a unit vector, so its magnitude is one.

So by applying this result to the two vectors ๐€ and ๐, well, we note that weโ€™re told ๐œƒ is the measure of the angle between the two vectors. And ๐ฐ is a unit vector, so its magnitude is one. We get the magnitude of two times the magnitude of ๐€ multiplied by the magnitude of ๐ times the magnitude of the sin of ๐œƒ.

And now we can simplify this even further. First, the magnitude of two is just equal to two. Next, ๐€ and ๐ are both unit vectors, so their magnitude is one. Finally, ๐œƒ is the measure of the angle between the two vectors. And when we see the angle, we mean the smaller of the two angles. In particular, this tells us that ๐œƒ is between zero and 180 degrees. And the sine of an angle between zero and 180 degrees is nonnegative. So the magnitude of sin of ๐œƒ in this case is just equal to sin ๐œƒ. This leaves us with two sin ๐œƒ, which is our final answer.

Therefore, we were able to show if ๐€ and ๐ are unit vectors and ๐œƒ is the measure of the angle between them, then the magnitude of the cross product between the difference between vectors ๐€ and ๐ and the sum of vectors ๐€ and ๐ is two sin ๐œƒ.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.